Re: gcd and lcm

It certainly seems valid, but it also seems too complex. Maybe it's just me, but I hate dealing with primes when I don't have to.

So I went back, and one of my methods was to show that lcm(a, b) = ab / gcd(a, b), which would suffice for this proof. But in my work, I went in circles because I didn't realize a little trick. So try this:

Let l = lcm(a, b) and g = gcd(a, b). Also, let n = ab / g for some integer n, since g must divide both a and b. What we wish to show is that n = l.

Since g = gcd(a, b), it stands that g | a and g | b. So let a = g*i and b = g*j for some integers i and j. Since a | l and b | l, it must be that ga'b' | l. But notice that n = ab / g = ga'gb' / g = ga'b'. So n | l and so n <= l.

Also, since g | a and g | b, it must be that (a/g) and (b/g) are integers. Thus, n = a(b/g) and n = (a/g)b, so a | n and b | n. So n is a multiple of a and b. Since l is the least such multiple, it must be that l <= n.

So l = n.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."